GATE EC · EE · IN

Network theorems

Network theorems let you replace all or part of a linear circuit with something simpler. Superposition splits the problem source by source, Thevenin and Norton reduce a network to one source and one impedance, and maximum power transfer then gives the best load in one line. Reciprocity, Millman, Miller, substitution and compensation are further shortcuts, and Tellegen's theorem is a power balance that holds for any lumped network. Common mistakes come from ignoring a theorem's conditions: switching off a dependent source, adding powers, using $R_{L} = R_{t h}$ for a resistive AC load, or applying reciprocity to a network that contains controlled sources.

16 previous-year questions · 16 solved · 5 practice questions

Linearity and superposition

A circuit element is linear if its voltage–current relation satisfies two rules. Homogeneity: scaling the cause scales the effect by the same factor. Additivity: the effect of a sum of causes is the sum of the separate effects. Resistors, inductors, capacitors and coupled coils with constant values are linear, and so are linear dependent sources (output proportional to one voltage or current). A circuit built only from linear elements and independent sources is a linear circuit. In a resistive circuit, and in AC steady state at a single frequency, every voltage and current is a weighted sum of the independent source values. In a transient problem, the initial capacitor voltages and inductor currents act as extra sources.

LINEARITY (SUPERPOSITION)

y = k_{1} x_{1} + k_{2} x_{2} + \dots + k_{n} x_{n}

$x_{i}$ = the independent source values; $y$ = any branch voltage or current. This holds for resistive circuits, and for AC steady state at one frequency, where $x_{i}$ and $y$ are phasors and the $k_{i}$ are complex numbers fixed by the circuit and the frequency. Dependent sources belong to the circuit and are never among the $x_{i}$ .

Superposition, step by step

Keep one independent source active. Set every other independent source to zero: an ideal voltage source becomes a short circuit and an ideal current source becomes an open circuit. Any internal resistance of a source stays in place.
Leave every dependent source in the circuit, still controlled by its own voltage or current. Dependent sources are never switched off.
Find the required voltage or current, using the same reference direction in every sub-circuit.
Repeat for each independent source (or for any convenient group of sources).
Add the partial responses algebraically. If the sources have different frequencies, add them as time functions. Compute power from the total voltage or current; the one exception, average power from components at distinct frequencies, is covered below.

Linearity gives two more shortcuts. If every independent source is multiplied by $k$ , every response is multiplied by $k$ . This is the basis of the unit-output method for ladders with one source: assume $1$ V or $1$ A at the far end, work back to the source, then scale. And if a response has been measured at as many independent source settings as there are unknown coefficients, the $k_{i}$ can be found without ever seeing the circuit.

EXAM TIP

Black-box linearity. A linear circuit has a voltage source $V_{s}$ and a current source $I_{s}$ , so any response is $y = a V_{s} + b I_{s}$ . Suppose $V_{s} = 10$ V, $I_{s} = 2$ A gives $y = 8$ V, and $V_{s} = 20$ V, $I_{s} = 1$ A gives $y = 13$ V. Then $10 a + 2 b = 8$ and $20 a + b = 13$ , so $a = 0.6$ and $b = 1 Ω$ . If the box also holds a fixed internal source that is never varied, add a constant term: $y = a V_{s} + b I_{s} + c$ . You then need three independent measurements.

EXAM TIP

Power does not superpose. Suppose one source alone drives $2$ A through a $10 Ω$ resistor and the other alone drives $3$ A in the same direction. Together they drive $5$ A, so $P = 25 \times 10 = 250$ W, not $40 + 90 = 130$ W. The missing part is the cross term $2 I_{1} I_{2} R$ . Superpose currents and voltages first, then compute power.

POWER AND RMS VALUE WITH COMPONENTS AT DIFFERENT FREQUENCIES

P = I_{0}^{2} R + k \sum I_{k}^{2} R, I_{rms} = I_{0}^{2} + k \sum I_{k}^{2}

$I_{0}$ = DC component; $I_{k}$ = rms value of the component at frequency $ω_{k}$ , with all the $ω_{k}$ non-zero and distinct. Here average powers do add, because the cross terms average to zero over time. Two components at the same frequency do not: combine them as phasors first. Example: $i = 3 + 42 sin ω t$ A gives $I_{rms} = 9 + 16 = 5$ A.

EXAM TIP

Sources at different frequencies (including DC). Solve each frequency in its own phasor circuit. For the DC source in steady state, an inductor is a short and a capacitor is an open. Convert each answer back to a time function, then add. Never add phasors that belong to different frequencies.

Thevenin and Norton equivalents

Seen from its terminals a–b, a linear two-terminal network behaves like one voltage source $V_{t h}$ in series with an impedance $Z_{t h}$ (Thevenin), with the $+$ terminal of $V_{t h}$ towards a. Equally, it behaves like one current source $I_{N}$ in parallel with the same impedance (Norton), with the arrow of $I_{N}$ pointing towards a. Only the network being replaced has to be linear; the load across a–b can be nonlinear, provided nothing couples the two sides except the two terminals (see the tip below). In AC, $V_{t h}$ and $Z_{t h}$ belong to one frequency; sources at several frequencies need a separate equivalent at each frequency.

$V_{t h}$ is the open-circuit voltage $V_{oc} = V_{a} - V_{b}$ . $I_{N}$ is the short-circuit current $I_{sc}$ , taken as flowing from a to b through the short. With these reference directions $V_{oc} = I_{sc} Z_{t h}$ , and source transformation converts one form into the other. For resistive DC circuits, write $R_{t h}$ for $Z_{t h}$ .

THEVENIN–NORTON RELATIONS

V_{t h} = V_{oc}, I_{N} = I_{sc} Z_{t h} = Z_{N} = \frac{V _{oc}}{I _{sc}}, V_{t h} = I_{N} Z_{t h}

$V_{oc} = V_{a} - V_{b}$ with a–b open; $I_{sc}$ flows from a to b through a short across a–b. The ratio cannot be used when both are zero. That is always the case for a network with no independent source, and it can also happen when the sources inside produce nothing at a–b (a balanced bridge, for example). Then use the test-source method.

SOURCE TRANSFORMATION

V_{s} in series with Z \leftrightarrow I_{s} = \frac{V _{s}}{Z} in parallel with Z

The arrow of $I_{s}$ points towards the terminal on the $+$ side of $V_{s}$ . The two forms are equivalent only as seen from the terminals: the current and power inside $Z$ differ. A dependent source can be transformed the same way, provided its controlling variable is not inside the $Z$ being moved. An ideal voltage source ( $Z = 0$ ) or an ideal current source ( $Z \to \infty$ ) cannot be transformed.

TEST-SOURCE METHOD

Z_{t h} = \frac{V _{t}}{I _{t}}_{independent sources off}

Short the independent voltage sources, open the independent current sources, and leave every dependent source active. Apply $V_{t}$ across a–b ( $+$ at a) and find the current $I_{t}$ entering terminal a from the test source (or inject $I_{t}$ into a and find $V_{t}$ ). This works whenever the equivalent exists, and it is the method to use when the network has no independent source.

Three ways to find the Thevenin resistance

Only independent sources in the network: switch them off and reduce by series–parallel combination.
Any linear network: compute $V_{oc}$ and $I_{sc}$ separately and take $R_{t h} = V_{oc} / I_{sc}$ . This works with dependent sources, provided $V_{oc}$ and $I_{sc}$ are not both zero.
Dependent sources present (always valid): switch off the independent sources only and apply a test source. A network with only dependent sources and resistors has $V_{t h} = 0$ and $I_{N} = 0$ , and its equivalent is just $R_{t h}$ .

WORKED EXAMPLE · EXAMPLE 1: THEVENIN EQUIVALENT WITH A DEPENDENT SOURCE

A $40$ V independent source (positive terminal up, negative terminal on the reference node b) is connected through a $10 Ω$ resistor to node a. Let $I_{x}$ be the current in the $10 Ω$ resistor, flowing from the source towards a. Two elements connect node a to b: a $20 Ω$ resistor, and a current-controlled current source of value $0.5 I_{x}$ that pushes current from b into node a. A variable load $R_{L}$ is connected across a–b. Find the Thevenin equivalent at a–b, the value of $R_{L}$ for maximum power, and that maximum power.

Solution

Open-circuit voltage. With a–b open, KCL at a gives (current in) = (current out through $20 Ω$ ): $I_{x} + 0.5 I_{x} = V_{a} /20$ , where $I_{x} = (40 - V_{a}) /10$ .
So $1.5 (40 - V_{a}) /10 = V_{a} /20$ , which gives $3 (40 - V_{a}) = V_{a}$ and $V_{a} = 30$ V. Hence $V_{t h} = 30$ V.
Short-circuit current. With a–b shorted, $V_{a} = 0$ and the $20 Ω$ resistor carries nothing. $I_{x} = 40/10 = 4$ A, and the dependent source adds $0.5 \times 4 = 2$ A. So $I_{sc} = 4 + 2 = 6$ A, flowing from a to b through the short.
$R_{t h} = V_{oc} / I_{sc} = 30/6 = 5 Ω$ , and the Norton current is $I_{N} = 6$ A.
Check with a test source. Short the $40$ V source and apply $V_{t}$ at a, with $I_{t}$ entering a. Now $I_{x} = (0 - V_{t}) /10 = - V_{t} /10$ . KCL at a: $I_{t} + I_{x} + 0.5 I_{x} = V_{t} /20$ , so $I_{t} = V_{t} /20 + 0.15 V_{t} = 0.2 V_{t}$ . That gives $R_{t h} = 5 Ω$ , as before.
The naive answer $10 ∥ 20 = 6.67 Ω$ , found by deleting the dependent source, is wrong.
Maximum power: $R_{L} = R_{t h} = 5 Ω$ and $P_{ma x} = V_{t h}^{2} / (4 R_{t h}) = 900/20 = 45$ W.

Answer: $V_{t h} = 30$ V, $R_{t h} = 5 Ω$ ( $I_{N} = 6$ A); maximum power at $R_{L} = 5 Ω$ , $P_{ma x} = 45$ W.

EXAM TIP

A dependent source can make $R_{t h}$ smaller or larger than the passive resistors suggest, and even zero or negative. A negative $R_{t h}$ is a correct result, not an arithmetic slip, but the maximum power formulas below then do not apply: they need $R_{t h} > 0$ . If $R_{t h} = 0$ , the network has no Norton equivalent; if $R_{t h}$ is infinite, it has no Thevenin equivalent.

EXAM TIP

Only cut the circuit where each dependent source and its controlling voltage or current are on the same side, and where no mutual inductance links the two sides. If the controlling variable lies in the load, first rewrite it in terms of the terminal voltage or current.

Maximum power transfer

Suppose the source network is fixed, so its Thevenin equivalent is fixed, and the load can vary. The load power $P_{L} = V_{t h}^{2} R_{L} / (R_{t h} + R_{L})^{2}$ is zero at $R_{L} = 0$ and tends to zero as $R_{L} \to \infty$ , so it peaks somewhere in between. For $R_{t h} > 0$ , setting $d P_{L} / d R_{L} = 0$ gives $R_{L} = R_{t h}$ . Throughout this section, power means power absorbed by the load (passive sign convention). For AC, all phasors are rms values.

MAXIMUM POWER TRANSFER, RESISTIVE LOAD

R_{L} = R_{t h}, P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}} = \frac{I _{N}^{2} R _{t h}}{4}

$R_{t h} > 0$ is fixed and $R_{L}$ varies. If the network has dependent sources, use the $R_{t h}$ found with them active.

LOAD POWER FOR ANY LOAD RESISTANCE

P_{L} = \frac{V _{t h}^{2} R _{L}}{( R _{t h} + R _{L} ) ^{2}}

$P_{L}$ rises for $R_{L} < R_{t h}$ and falls for $R_{L} > R_{t h}$ . So if $R_{L}$ is restricted to a range that excludes $R_{t h}$ , the best choice is the allowed value nearest $R_{t h}$ .

EFFICIENCY OF THE THEVENIN EQUIVALENT

η = \frac{R _{L}}{R _{L} + R _{t h}}, η = 50% at R_{L} = R_{t h}

This is the efficiency of the Thevenin equivalent circuit only. The power drawn from the sources inside the original network is different, so the real efficiency at matching is generally not 50%.

EXAM TIP

Maximum power transfer is not maximum efficiency. As $R_{L} \to \infty$ the efficiency approaches 100%, but the load power falls towards zero. The reverse problem has a different answer: if $R_{L}$ is fixed and the source resistance can be chosen, the load power is largest when the source resistance is $0$ , not when it equals $R_{L}$ .

For AC, the source is the rms phasor $V_{t h}$ behind $Z_{t h} = R_{t h} + j X_{t h}$ , and the load is $Z_{L} = R_{L} + j X_{L}$ . The average power in the load is $P_{L} = ∣ V_{t h} ∣^{2} R_{L} / [(R_{t h} + R_{L})^{2} + (X_{t h} + X_{L})^{2}]$ . Only $R_{L}$ absorbs average power. Which load is best depends on which parts of $Z_{L}$ are free to vary.

AC MAXIMUM POWER TRANSFER, CONJUGATE MATCH

Z_{L} = Z_{t h}^{*} = R_{t h} - j X_{t h}, P_{ma x} = \frac{∣ V _{t h} ∣ ^{2}}{4 R _{t h}}

Use this when both $R_{L}$ and $X_{L}$ are free (and $R_{t h} > 0$ ). The reactances cancel, so the loop is resonant. $∣ V_{t h} ∣$ is an rms value; if the phasor is a peak value $V_{m}$ , then $P_{ma x} = ∣ V_{m} ∣^{2} / (8 R_{t h})$ .

AC SOURCE, PURELY RESISTIVE LOAD

R_{L} = ∣ Z_{t h} ∣ = R_{t h}^{2} + X_{t h}^{2} P_{ma x} = \frac{∣ V _{t h} ∣ ^{2}}{2 ( R _{t h} + ∣ Z _{t h} ∣ )}

Use this when the load must be a pure resistor ( $X_{L} = 0$ ). Choosing $R_{L} = R_{t h}$ here is a common error.

FIXED LOAD REACTANCE, VARIABLE LOAD RESISTANCE

R_{L} = R_{t h}^{2} + (X_{t h} + X_{L})^{2} P_{ma x} = \frac{∣ V _{t h} ∣ ^{2}}{2 ( R _{t h} + R _{L} )}

$X_{L}$ is fixed and $R_{L}$ varies; $P_{ma x}$ is evaluated at that $R_{L}$ . The pure-resistor case is $X_{L} = 0$ , which gives back $R_{L} = ∣ Z_{t h} ∣$ .

FIXED LOAD ANGLE, VARIABLE MAGNITUDE

∣ Z_{L} ∣ = ∣ Z_{t h} ∣ P_{ma x} = \frac{∣ V _{t h} ∣ ^{2} cos θ _{L}}{2∣ Z _{t h} ∣ [ 1 + cos ( θ _{t h} - θ _{L} ) ]}

Use this for a load $Z_{L} = ∣ Z_{L} ∣∠ θ_{L}$ whose angle $θ_{L}$ is fixed (a fixed power factor) and whose magnitude alone varies; $θ_{t h}$ is the angle of $Z_{t h}$ . Setting $θ_{L} = 0$ gives back the pure-resistor result, and $θ_{L} = - θ_{t h}$ gives back the conjugate match.

Which AC load gives maximum power

What the load can vary	Condition for maximum power	Maximum power (rms $V_{t h}$ )
$R_{L}$ and $X_{L}$ freely	$Z_{L} = Z_{t h}^{*}$	$\frac{∣ V _{t h} ∣ ^{2}}{4 R _{t h}}$
$R_{L}$ only, pure resistor	$R_{L} = ∣ Z_{t h} ∣$	$\frac{∣ V _{t h} ∣ ^{2}}{2 ( R _{t h} + ∣ Z _{t h} ∣)}$
$R_{L}$ only, $X_{L}$ fixed	$R_{L} = R_{t h}^{2} + (X_{t h} + X_{L})^{2}$	$\frac{∣ V _{t h} ∣ ^{2}}{2 ( R _{t h} + R _{L} )}$ at that $R_{L}$
$X_{L}$ only, $R_{L}$ fixed	$X_{L} = - X_{t h}$	$\frac{∣ V _{t h} ∣ ^{2} R _{L}}{( R _{t h} + R _{L} ) ^{2}}$
$∣ Z_{L} ∣$ only, angle $θ_{L}$ fixed	$∣ Z_{L} ∣ = ∣ Z_{t h} ∣$	$\frac{∣ V _{t h} ∣ ^{2} cos θ _{L}}{2 ∣ Z _{t h} ∣ [ 1 + cos ( θ _{t h} - θ _{L} ) ]}$

WORKED EXAMPLE · EXAMPLE 2: BEST AC LOAD, FREE AND RESISTIVE

At the operating frequency, the Thevenin equivalent of a network at its load terminals is $V_{t h} = 20∠ 0^{\circ}$ V (rms) in series with $Z_{t h} = 3 + j 4 Ω$ . Find the maximum average power the load can absorb (a) if the load impedance is unrestricted, and (b) if the load must be a pure resistor. (c) Show that a $3 Ω$ resistor does worse than the answer to (b).

Solution

(a) Conjugate match: $Z_{L} = 3 - j 4 Ω$ . The loop impedance is $6 Ω$ , so $∣ I ∣ = 20/6$ A rms and $P = ∣ I ∣^{2} \times 3 = 400/12 = 33.33$ W. The formula $∣ V_{t h} ∣^{2} / (4 R_{t h})$ gives the same result.
(b) Pure resistor: $R_{L} = ∣ Z_{t h} ∣ = 9 + 16 = 5 Ω$ . The loop impedance is $8 + j 4 Ω$ , so $∣ I ∣^{2} = 400/ (64 + 16) = 5 A^{2}$ and $P = 5 \times 5 = 25$ W.
Shortcut check for (b): $∣ V_{t h} ∣^{2} / (2 (R_{t h} + ∣ Z_{t h} ∣)) = 400/ (2 \times 8) = 25$ W.
(c) With $R_{L} = 3 Ω$ : $∣ I ∣^{2} = 400/ (36 + 16) = 7.69 A^{2}$ , so $P = 7.69 \times 3 = 23.08$ W, which is less than $25$ W.

Answer: (a) $33.33$ W with $Z_{L} = 3 - j 4 Ω$ ; (b) $25$ W with $R_{L} = 5 Ω$ ; (c) $23.08$ W with $R_{L} = 3 Ω$ .

Reciprocity, Tellegen and Millman

A network is reciprocal if the ratio of a response to its excitation stays the same when the excitation and the response swap places. Only two swaps are guaranteed. The first is an ideal voltage source with the short-circuit current in another branch (an ideal ammeter). The second is an ideal current source with the open-circuit voltage across another pair of terminals (an ideal voltmeter). When a voltage source moves, the branch it leaves is short-circuited. When a current source moves, the branch it leaves is left open.

The two forms of reciprocity

Excitation	Response	Quantity unchanged by the swap
Ideal voltage source $V$ in branch 1	Short-circuit current $I$ in branch 2	Transfer admittance $I / V$
Ideal current source $I$ at port 1	Open-circuit voltage $V$ at port 2	Transfer impedance $V / I$

Conditions for reciprocity

All elements linear and time-invariant.
All elements bilateral. Resistors, inductors, capacitors, coupled coils ( $M_{12} = M_{21}$ ) and ideal transformers qualify; diodes do not.
No dependent sources. A controlled source, a transistor model or an op-amp generally makes the network non-reciprocal.
A single independent source, with all initial conditions zero.
The ratio must pair a voltage source with a short-circuit current, or a current source with an open-circuit voltage.

RECIPROCITY CONDITIONS FOR A TWO-PORT

z_{12} = z_{21}, y_{12} = y_{21}, A D - B C = 1 h_{12} = - h_{21}, g_{12} = - g_{21}

Convention: $I_{1}$ and $I_{2}$ both flow INTO their ports, and ABCD is defined by $V_{1} = A V_{2} - B I_{2}$ , $I_{1} = C V_{2} - D I_{2}$ . Where the parameter sets exist, any one condition implies the others. Symmetry ( $z_{11} = z_{22}$ , $A = D$ ) is a separate, stronger property.

EXAM TIP

In a reciprocal network, suppose $20$ V at port 1 drives $1.5$ A through a short at port 2. Then $5$ V at port 2 drives $1.5 \times 5/20 = 0.375$ A through a short at port 1: reciprocity plus linearity. Voltage ratios do not swap like this. $V_{2} / V_{1}$ with port 2 open equals $z_{21} / z_{11}$ , while the reverse ratio equals $z_{12} / z_{22}$ . The two are equal only if the network is also symmetric.

Tellegen's theorem uses only KCL and KVL on the network graph. It therefore holds for any lumped network: linear or nonlinear, time-varying or not, with any sources. Take associated reference directions (passive sign convention) on every branch, so that $v_{k} i_{k}$ is the power absorbed by branch $k$ .

TELLEGEN'S THEOREM

k = 1 \sum b v_{k} i_{k} = 0, k = 1 \sum b v_{k}^{'} i_{k}^{''} = 0

The sums run over all $b$ branches. The first form is conservation of power: the power delivered by the sources equals the power absorbed by everything else. The second form pairs voltages from one network with currents from another that has the same graph and the same reference directions. It also holds for one network at two different instants.

Millman's theorem combines several parallel branches between two nodes a and b. Each branch is a voltage source $V_{k}$ (positive terminal towards a) in series with an impedance $Z_{k}$ , with admittance $Y_{k} = 1/ Z_{k}$ . The theorem is nodal analysis at node a written in one line. The result is the Thevenin equivalent of the whole parallel group.

MILLMAN'S THEOREM

V_{ab} = \frac{\sum _{k} V _{k} Y _{k} + \sum _{j} I _{j}}{\sum _{k} Y _{k}}, Z_{e q} = \frac{1}{\sum _{k} Y _{k}}

$I_{j}$ = ideal current sources between the two nodes, directed into a; they add nothing to the denominator. A passive branch enters with $V_{k} = 0$ . An ideal voltage source with $Z_{k} = 0$ cannot be included, because it fixes $V_{ab}$ directly.

Miller, substitution and compensation theorems

Miller's theorem removes an impedance $Z$ that bridges two nodes 1 and 2, typically the input and output of an amplifier. It requires the node voltages, measured from ground, to satisfy $V_{2} = K V_{1}$ . The current leaving node 1 through $Z$ is $(V_{1} - V_{2}) / Z = V_{1} (1 - K) / Z$ . An impedance $Z / (1 - K)$ from node 1 to ground would draw exactly this current. The same argument at node 2 gives the second impedance.

MILLER'S THEOREM

Z_{1} = \frac{Z}{1 - K}, Z_{2} = \frac{Z}{1 - 1/ K} = \frac{Z K}{K - 1} K = \frac{V _{2}}{V _{1}}

$Z$ is removed and replaced by $Z_{1}$ from node 1 to ground and $Z_{2}$ from node 2 to ground. $K$ must be the gain with $Z$ connected. For $K > 1$ , $Z_{1}$ is negative; for $0 < K < 1$ (a follower), $Z_{2}$ is negative.

MILLER CAPACITANCE

C_{1} = C (1 - K), C_{2} = C (1 - \frac{1}{K})

This is the capacitor version (impedance $1/ (j ω C)$ ). For an inverting stage, $K = - A$ with $A > 0$ . Then $C_{1} = C (1 + A)$ is large (the Miller effect), and $C_{2} \approx C$ when $A ≫ 1$ . A bridging resistor $R$ becomes $R / (1 - K)$ at the input.

EXAM TIP

The Miller split reproduces the node voltages only for the $K$ that holds with $Z$ in place. It therefore cannot be used to find the output impedance of the stage. Using a gain computed without $Z$ connected is an approximation.

Substitution theorem. Suppose a branch carries current $I$ and has voltage $V$ across it. It can be replaced by an independent voltage source $V$ (same polarity) or an independent current source $I$ (same direction), and every other voltage and current stays the same. The only condition is that the new network still has a unique solution. This theorem justifies treating a known node voltage or branch current as a source when a circuit is analysed in pieces.

Compensation theorem. In a linear network, let one branch with impedance $Z$ carry current $I$ , and let its impedance change to $Z + Δ Z$ . Switch off all independent sources and insert a single voltage source $V_{c} = I Δ Z$ in series with the modified branch, with its polarity opposing $I$ . The currents this source produces are exactly the changes in every branch current.

COMPENSATION THEOREM

V_{c} = I Δ Z, Δ I_{branch} = - \frac{I Δ Z}{Z _{t h} + Z + Δ Z}

$I$ = original branch current. $Z_{t h}$ = impedance seen by the branch looking into the rest of the network, with independent sources off. The minus sign says that increasing the impedance reduces the branch current. A typical use is the error caused by inserting an ammeter of resistance $Δ Z$ .

Choosing a theorem and checking its conditions

Where each theorem applies

Theorem	Applies to	Watch out for
Superposition	Linear circuits with any number of independent sources; dependent sources allowed	Dependent sources stay on; power is not superposable; different frequencies add in the time domain
Thevenin / Norton	A linear network seen from two terminals; the load may be nonlinear	A dependent source and its controlling variable on the same side of the cut, and no mutual inductance across it; $I_{sc}$ direction from a to b
Maximum power transfer	Fixed source network with $R_{t h} > 0$ , variable load	Conjugate match only if $X_{L}$ is free; pure resistor needs $R_{L} = ∣ Z_{t h} ∣$ ; 50% efficiency holds only for the Thevenin model
Reciprocity	Linear, bilateral, time-invariant networks with no dependent sources and a single source	Only the voltage source with short-circuit current and current source with open-circuit voltage swaps
Tellegen	Any lumped network, linear or not	Associated reference directions on every branch; the two-network form needs the same graph
Millman	Parallel branches of sources with series impedances	A branch that is an ideal voltage source ( $Z_{k} = 0$ ) cannot enter the formula
Miller	An impedance between two nodes with a known ratio $V_{2} / V_{1}$	$K$ taken with $Z$ connected; not valid for output impedance
Substitution	Any network with a unique solution	The replacement source must equal the branch's actual voltage or current
Compensation	A linear network in which one branch impedance changes	The changes are found with independent sources off and the modified impedance in place

EXAM TIP

Quick self-checks. With a–b open, the Thevenin model should give back $V_{oc}$ ; with a–b shorted, it should give back $I_{sc}$ . At a matched resistive load the load voltage is $V_{t h} /2$ and the load current is $I_{N} /2$ : solving the original circuit with $R_{L} = R_{t h}$ connected should give the same values. If it does not, recheck $R_{t h}$ .

PREVIOUS-YEAR QUESTIONS

16 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2024 · GATE ECE 2022 · GATE ECE 2020 · GATE EE 2026 · GATE EE 2023 · GATE EE 2020 · GATE IN 2025 · GATE IN 2023 · GATE IN 2022 · GATE IN 2020 · GATE IN 2018

GATE 2026 EE Q.24

GATE 2026 · EEQ.24 · 1 mark · MCQ

For the circuit shown, which one of the following options correctly identifies the Thevenin's equivalent parameters between nodes Y and Z?

GATE 2026 EE question 24: Four nodes: X (left), Y (top middle), Z (top right), W (bottom middle). — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$V_{T H} = 100 V$ , $R_{T H} = 10 k Ω$
(B)
$V_{T H} = 140 V$ , $R_{T H} = 0 Ω$
(C)
$V_{T H} = 100 V$ , $R_{T H} = 0 Ω$
(D)
$V_{T H} = 140 V$ , $R_{T H} = 10 k Ω$

Answer and solution

GATE 2026 EE Q.24: answer and solution

GATE 2026 EE Q.44

GATE 2026 · EEQ.44 · 2 marks · MCQ

The terminal voltage and current of a linear electrical network shown in Figure (a) are given in the table.

| Terminal voltage ( $v_{t}$ ) | Terminal current ( $i_{t}$ ) |
|---|---|
| 18 V | $- 0.5$ A |
| 30 V | 0.5 A |
| 36 V | 1.0 A |

The correct choice for the parameters ( $I_{N}$ , $R_{N}$ ) of the Norton equivalent circuit shown in Figure (b) is:

GATE 2026 EE question 44: Figure (a): a box labelled 'Electrical Network' with two terminals brought out to the right. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$I_{N} = 3.0 A$ , $R_{N} = 24.0 Ω$
(B)
$I_{N} = 12.0 A$ , $R_{N} = 2.0 Ω$
(C)
$I_{N} = 2.0 A$ , $R_{N} = 12.0 Ω$
(D)
$I_{N} = 2.0 A$ , $R_{N} = 24.0 Ω$

Answer and solution

GATE 2026 EE Q.44: answer and solution

GATE 2026 EE Q.60

GATE 2026 · EEQ.60 · 2 marks · NAT

Consider the two-port network shown. For maximum power transfer to the resistive load ( $R_{L}$ ), the value of $R_{L}$ should be ________ $Ω$ (Round off to two decimal places)

GATE 2026 EE question 60: An ideal 10 V DC voltage source, + terminal on top, drives port 1; — Figure from the GATE 2026 EE paper. It is part of the question.

Answer and solution

GATE 2026 EE Q.60: answer and solution

GATE 2025 IN Q.45

GATE 2025 · INQ.45 · 2 marks · MCQ

The bridge circuit, shown in Figure (a), can be equivalently represented using the circuit shown in Figure (b). The values of $R_{1}$ , $R_{2}$ , and $V_{C}$ in the equivalent circuit are

GATE 2025 IN question 45: Figure (a): A diamond-shaped bridge. — Figure from the GATE 2025 IN paper. It is part of the question.

(A)
$R_{1} = 6 k Ω$ , $R_{2} = 3 k Ω$ , and $V_{C} = 9 V$
(B)
$R_{1} = 3 k Ω$ , $R_{2} = 6 k Ω$ , and $V_{C} = 4.5 V$
(C)
$R_{1} = 2 k Ω$ , $R_{2} = 2 k Ω$ , and $V_{C} = 9 V$
(D)
$R_{1} = 2 k Ω$ , $R_{2} = 2 k Ω$ , and $V_{C} = 4.5 V$

Answer and solution

GATE 2025 IN Q.45: answer and solution

GATE 2025 IN Q.61

GATE 2025 · INQ.61 · 2 marks · NAT

In the circuit shown, the galvanometer (G) has an internal resistance of 100 Ω. The galvanometer current $I_{G}$ is ________ μA (rounded off to the nearest integer).

GATE 2025 IN question 61: A Wheatstone bridge drawn as a diamond. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

GATE 2025 IN Q.61: answer and solution

GATE 2024 EC Q.62

GATE 2024 · ECQ.62 · 2 marks · NAT

In the network shown below, maximum power is to be transferred to the load $R_{L}$ .

[Figure]

The value of $R_{L}$ (in $Ω$ ) is ________ .

GATE 2024 EC question 62: A 50 V independent DC source sits in the left vertical branch, + terminal at the top. — Figure from the GATE 2024 EC paper. It is part of the question.

Answer and solution

GATE 2024 EC Q.62: answer and solution

GATE 2023 EE Q.51

GATE 2023 · EEQ.51 · 2 marks · NAT

For the circuit shown, if $i = sin 1000 t$ , the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminals $a$ - $b$ at time $t = 5$ ms is ______ (Round off to 2 decimal places).

GATE 2023 EE question 51: Three parallel paths between a top rail and a bottom rail. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

GATE 2023 EE Q.51: answer and solution

GATE 2023 IN Q.31

GATE 2023 · INQ.31 · 1 mark · NAT

In the circuit given, the Thevenin equivalent resistance $R_{t h}$ across the terminals 'a' and 'b' is ________ $Ω$ (rounded off to one decimal place).

GATE 2023 IN question 31: Planar ladder circuit with a common bottom rail that is terminal b. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

GATE 2023 IN Q.31: answer and solution

GATE 2022 EC Q.54

GATE 2022 · ECQ.54 · 2 marks · NAT

A linear 2-port network is shown in Fig. (a). An ideal DC voltage source of 10 V is connected across Port 1. A variable resistance $R$ is connected across Port 2. As $R$ is varied, the measured voltage and current at Port 2 is shown in Fig. (b) as a $V_{2}$ versus $- I_{2}$ plot. Note that for $V_{2} = 5 V$ , $I_{2} = 0 mA$ , and for $V_{2} = 4 V$ , $I_{2} = - 4 mA$ .

When the variable resistance $R$ at Port 2 is replaced by the load shown in Fig. (c), the current $I_{2}$ is ________ mA (rounded off to one decimal place).

GATE 2022 EC question 54: Fig. — Figure from the GATE 2022 EC paper. It is part of the question.

Answer and solution

GATE 2022 EC Q.54: answer and solution

GATE 2022 IN Q.42

GATE 2022 · INQ.42 · 2 marks · MCQ

A resistor ladder digital-to-analog converter (DAC) receives a digital input that results in the circuit having the state as shown in the figure. For this digital input, the Thevenin voltage, $V_{t h}$ , and Thevenin resistance, $R_{t h}$ , as seen at the output node are ________

GATE 2022 IN question 42: R-2R ladder. — Figure from the GATE 2022 IN paper. It is part of the question.

(A)
$V_{t h} = 0.5 V$ , $R_{t h} = 1 k Ω$
(B)
$V_{t h} = 0.5 V$ , $R_{t h} = 2 k Ω$
(C)
$V_{t h} = 1 V$ , $R_{t h} = 1 k Ω$
(D)
$V_{t h} = 1 V$ , $R_{t h} = 2 k Ω$

Answer and solution

GATE 2022 IN Q.42: answer and solution

GATE 2022 IN Q.54

GATE 2022 · INQ.54 · 2 marks · NAT

Given Circuit A with currents $I_{1}$ and $I_{2}$ as shown, the current $I_{3}$ in Circuit B (in amperes), is ________ (round off to one decimal place)

GATE 2022 IN question 54: Two separate ladder circuits, A and B, each with its own bottom rail (the reference node). — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

GATE 2022 IN Q.54: answer and solution

GATE 2020 EC Q.9

GATE 2020 · ECQ.9 · 1 mark · MCQ

In the circuit shown below, the Thevenin voltage $V_{T H}$ is

GATE 2020 EC question 9: A ladder circuit with a common bottom (reference) rail. — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
2.4 V
(B)
2.8 V
(C)
3.6 V
(D)
4.5 V

Answer and solution

GATE 2020 EC Q.9: answer and solution

GATE 2020 EE Q.19

GATE 2020 · EEQ.19 · 1 mark · NAT

The Thevenin equivalent voltage, $V_{T H}$ , in V (rounded off to 2 decimal places) of the network shown below, is ________

GATE 2020 EE question 19: The circuit has a common bottom (reference) rail and a top rail, reading left to right. — Figure from the GATE 2020 EE paper. It is part of the question.

Answer and solution

GATE 2020 EE Q.19: answer and solution

GATE 2020 IN Q.49

GATE 2020 · INQ.49 · 2 marks · NAT

A circuit consisting of capacitors, DC voltage source and an amplifier having a voltage gain G = −5 is shown in the figure. The effective capacitance across the nodes A and B (in µF, rounded off to one decimal place) is ____

GATE 2020 IN question 49: Ladder circuit with an amplifier. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

GATE 2020 IN Q.49: answer and solution

GATE 2018 IN Q.7

GATE 2018 · INQ.7 · 1 mark · MCQ

The Thevenin equivalent circuit representation across terminals p-q of the circuit shown in the figure is a

GATE 2018 IN question 7: A 4 V independent voltage source sits on the left, + terminal at the top and - terminal at the bottom. — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
1 V source in series with 150 kΩ
(B)
1 V source in parallel with 100 kΩ
(C)
2 V source in series with 150 kΩ
(D)
2 V source in parallel with 200 kΩ

Answer and solution

GATE 2018 IN Q.7: answer and solution

GATE 2018 IN Q.30

GATE 2018 · INQ.30 · 2 marks · NAT

In the given circuit, superposition is applied. When $V_{2}$ is set to 0 V, the current $I_{2}$ is $- 6$ A. When $V_{1}$ is set to 0 V, the current $I_{1}$ is $+ 6$ A. Current $I_{3}$ (in A) when both sources are applied will be (up to two decimal places) ______.

GATE 2018 IN question 30: Two-mesh circuit with a common bottom (reference) wire. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

GATE 2018 IN Q.30: answer and solution

PRACTICE

5 questions to test yourself

PRACTICE 11 mark · MCQ

A linear circuit contains two independent sources, A and B. With only A active (B set to zero), a $10 Ω$ resistor R carries $2$ A. With only B active (A set to zero), R carries $3$ A in the same direction. With both sources active, the power dissipated in R is

(A)
$130$ W
(B)
$250$ W
(C)
$10$ W
(D)
$50$ W

Answer and solution

Answer: (B)

By superposition, the currents add: $I = 2 + 3 = 5$ A.
$P = I^{2} R = 25 \times 10 = 250$ W.
Option A adds the individual powers ( $40 + 90$ ) and so drops the cross term $2 I_{A} I_{B} R = 120$ W. Option C would apply if the currents were opposite ( $1$ A).

PRACTICE 22 marks · NAT

A $10$ V independent source (positive terminal up, negative terminal on the reference node) is connected through a $4 Ω$ resistor to node x. Let $I_{x}$ be the current in the $4 Ω$ resistor, flowing from the source towards x. A series branch connects node x to the reference node: a $2 Ω$ resistor attached to x, followed by a current-controlled voltage source of value $4 I_{x}$ volts. The positive terminal of this source is at the resistor end and its negative terminal is at the reference node. A variable resistor $R_{L}$ is connected between x and the reference node. The maximum power (in W) that can be delivered to $R_{L}$ is ______.

Answer and solution

Answer: $11.25$ (accepted range $\pm 0.05$ )

Open circuit. The branch current flowing down from x is $(V_{x} - 4 I_{x}) /2$ , and KCL at x makes it equal to $I_{x}$ . So $V_{x} = 6 I_{x}$ . With $I_{x} = (10 - V_{x}) /4$ this gives $I_{x} = 1$ A and $V_{oc} = 6$ V.
Short circuit ( $V_{x} = 0$ ): $I_{x} = 10/4 = 2.5$ A. The branch current flowing down is $(0 - 10) /2 = - 5$ A, meaning $5$ A flows up into x. So $I_{sc} = 2.5 + 5 = 7.5$ A.
$R_{t h} = 6/7.5 = 0.8 Ω$ . Test-source check (source shorted, $V_{t}$ at x): $I_{x} = - V_{t} /4$ and the downward branch current is $(V_{t} + V_{t}) /2 = V_{t}$ . So $I_{t} = V_{t} /4 + V_{t} = 1.25 V_{t}$ , confirming $R_{t h} = 0.8 Ω$ .
$P_{ma x} = V_{t h}^{2} / (4 R_{t h}) = 36/3.2 = 11.25$ W.

PRACTICE 32 marks · MSQ

Which of the following statements is/are correct?

(A)
Tellegen's theorem, $\sum_{k} v_{k} i_{k} = 0$ , holds for a network containing nonlinear and time-varying elements, provided the branch voltages satisfy KVL, the branch currents satisfy KCL, and every branch uses associated reference directions.
(B)
In a linear network containing dependent sources, the Thevenin resistance seen at a pair of terminals can be negative.
(C)
While applying superposition, every dependent source is replaced by a short circuit (if it is a voltage source) or an open circuit (if it is a current source), along with the independent sources that are not acting.
(D)
Whenever a load resistor equals the Thevenin resistance of the linear resistive network feeding it, exactly half of the total power supplied by the independent sources inside the network is delivered to the load.

Answer and solution

Answer: (A), (B)

A is true. Tellegen's theorem depends only on KCL, KVL and the graph, not on what the elements are.
B is true. Example: a $10 Ω$ resistor across a–b in parallel with a dependent source that injects twice the resistor current into a. A test voltage $V_{t}$ gives $I_{t} = V_{t} /10 - 2 V_{t} /10 = - V_{t} /10$ , so $R_{t h} = - 10 Ω$ .
C is false. Dependent sources are never switched off in superposition.
D is false. The 50% figure holds only for the Thevenin equivalent. Counter-example: a $10$ V source, $10 Ω$ in series and $10 Ω$ in shunt gives $R_{t h} = 5 Ω$ and $V_{t h} = 5$ V. With $R_{L} = 5 Ω$ , $P_{L} = 1.25$ W while the source delivers $10 \times 0.75 = 7.5$ W, so only $16.7%$ reaches the load.

PRACTICE 42 marks · MCQ

At the operating frequency, the Thevenin equivalent of a network at its load terminals is $50∠ 0^{\circ}$ V (rms) in series with $6 + j 8 Ω$ . The load is a pure resistor whose value is chosen to maximise the average power it absorbs. This maximum average power is

(A)
$104.17$ W
(B)
$78.13$ W
(C)
$72.12$ W
(D)
$125.00$ W

Answer and solution

Answer: (B)

For a pure-resistor load, $R_{L} = ∣ Z_{t h} ∣ = 36 + 64 = 10 Ω$ .
$∣ I ∣^{2} = 2500/ [(6 + 10)^{2} + 8^{2}] = 2500/320 = 7.8125 A^{2}$ , so $P = 7.8125 \times 10 = 78.125$ W.
Check: $∣ V_{t h} ∣^{2} / (2 (R_{t h} + ∣ Z_{t h} ∣)) = 2500/32 = 78.125$ W.
Option A, $2500/24$ , would need a conjugate-matched load, which is not allowed here. Option C is what the wrong choice $R_{L} = R_{t h} = 6 Ω$ gives: $2500 \times 6/ (1 2^{2} + 8^{2}) = 15000/208 = 72.12$ W, which is less than the maximum.

PRACTICE 51 mark · NAT

An inverting amplifier stage has voltage gain $V_{o} / V_{i} = - 49$ with the capacitor below connected. A $2$ pF capacitor is connected between its input and output nodes. Using Miller's theorem, the equivalent capacitance (in pF) from the input node to ground due to this capacitor is ______.

Answer and solution

Answer: $100$ (accepted range $\pm 0.5$ )

Miller input capacitance: $C_{1} = C (1 - K)$ with $K = - 49$ .
$C_{1} = 2 \times (1 + 49) = 100$ pF.
For reference, the output-side capacitance is $C_{2} = C (1 - 1/ K) = 2 \times (1 + 1/49) \approx 2.04$ pF.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

superposition, Thevenin's theorem, Norton’s theorem, reciprocity … maximum power transfer

GATE EE

Network Theorems: Thevenin’s, Norton’s, Superposition and Maximum Power Transfer theorem

GATE IN

superposition, Thevenin’s, Norton’s, Miller’s, maximum power transfer and reciprocity theorems

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